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प्रश्न
\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
विकल्प
−π
π
\[- \frac{1}{\pi}\]
\[\frac{1}{\pi}\]
उत्तर
−π
\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
\[ = \lim_{h \to 0} \frac{\sin \pi\left( 1 + h \right)}{\left( 1 + h \right) - 1}\]
\[ = \lim_{h \to 0} \frac{\sin \left( \pi + \pi h \right)}{h}\]
\[ = \lim_{h \to 0} \frac{- \sin \pi h}{h}\]
\[ = \lim_{h \to 0} - \left( \frac{\sin \pi h}{\pi h} \right)\pi\]
\[ = - \pi\]
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